I've solved currents earlier with loop-method, got 2.965 A for the current that goes through the resistor. Got 11.86 Volts as Vth and 1.842 Ohms as Rin (Possibly wrong..). However I'm unsure where to go from here.. 11.86V / 2.965 A = 4 Ohms.. like inner resistance (Rin) wouldn't matter?

I thought that calculating current over the resistor was simply Vth/Rin + original resistance... In which case the current would be 2A (11.86 / 5.842)

I admit that I'm quite terrible using Thevenins. It's hard to find good examples of it, or maybe I was taught it differently than it's presented in most places.. I understand how it's useful but haven't fully grasped how to use it. Maybe source of the problem is that we were taught it by using circuits where it really wasn't needed (solvable otherwise)

The assingment was just "Find the current of R4 (the 4 Ohm resistor between A and B) using Thevenin". I haven't been taught superpositioning, Just "base law method" (I1 = I2 + I3, then voltage equations) and loop method (1 equation / loop). So not sure what to do with it.

but for current produced by an individual source (either current or voltage source) u have to use superposition theorem

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Ah, I see. Just looked up the theorem on this site. That's pretty smart I wonder why we weren't taught that before..

I'll check it out tomorrow morning.

hgmjr said:

Step one of the process would be to derive the thevenin equivalent voltage and resistance associated with the 7 ohm and 10 ohm resistor and the 10V power source.

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I admit that I don't exactly understand that. I'm not native English speaker, and we weren't really taught Thevenin by terms & theory, just by examples in classes like "Here we have this 2-loop circuit. Now we snap out this resistor, then calculate voltage over it, which is same as Thevenin's voltage, then we determine the total resistance of the circuit, which will then be Thevenin's resistance et cetera et cetera". Which is the source of my problem in not understanding what to do The article here is very helpful though, I'll try to learn it by tomorrow (Have this exercise due Monday)

Thanks for replies to you both

edit: Oh, 1 thing. If it's not too much work, could anyone calculate the real Thevenin resistance and Thevenin voltage and post the values, In case I made mistakes. I'm partly doing it trial by error, so If I have wrong values to begin with it's going to be a problem. I still have to do all the calculations and whatever on paper, so I'm not asking anyone to solve it totally for me. Just some guidelines

Temporarily ignore all components and voltages sources and just concentrate on the 7 ohm resistor, 10 ohm resistor, and the 10V power source. What is the Thevenin's voltage and resistance for this simple network? If you will calculate this then I will provide you with the next step in the process.

Temporarily ignore all components and voltages sources and just concentrate on the 7 ohm resistor, 10 ohm resistor, and the 10V power source. What is the Thevenin's voltage and resistance for this simple network? If you will calculate this then I will provide you with the next step in the process.

hgmjr

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What is Thevenins voltage? I've been taught visually to figure out the voltage over two points (ab), f.x. a (load) resistor. That's all. However if we still take the rest of the circuit to consideration (It exists, that's all), then it would be V1R2/(R1+R2) = 100 / 17 = 5.89 V according to voltage divider?

Thevenin's Resistance would be 17 Ω, since both resistors are in serial.

You have calculated the correct Thevenin's voltage. You need to take another look at the Thevenin's resistance however.

hgmjr

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Alright. might R1R2/(R1+R2) be the correct formula for Thevenins Resistance then? Then it would be 4,12 Ω. That would be what the theory says when considering that there are other nodes connecting to the loop we're looking at, and following the same pattern as calculating Thevenin's voltage above.

You now have the ingredients with which to simplify your circuit into two loops.

You can redraw your circuit now to include the new voltage source and the new thevenin resistance together with the 5 ohm resistor in series with the thevenin's resistance. You will then have a two loop equation.

You can now perform the same Thevenin's calculation with the 90V, the 10 ohm, the 4 ohm and the 20V source to get the circuit further simplified.

You now have the ingredients with which to simplify your circuit into two loops.

You can redraw your circuit now to include the new voltage source and the new thevenin resistance together with the 5 ohm resistor in series with the thevenin's resistance. You will then have a two loop equation.

You can now perform the same Thevenin's calculation with the 90V, the 10 ohm, the 4 ohm and the 20V source to get the circuit further simplified.

hgmjr

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Hey, Thanks alot. I had a hunch that you were after something like this (Changing it to two loops), but didn't want to ask more stupid questions before figuring out more It should be quite simple now, I'll get back to it tomorrow morning and post the results (Almost Midnight where I live now).

Okay, I was too quick to celebrate. I attempted many calculations, and one of them might've even been correct, but I'm not sure... So I scrapped them. I'll have to take another look at this method of simplifying circuits (All hints, help and examples appreciated). I did, however, take another approach to this problem, and I think I managed to solve it.

It's originally 3-loop circuit. However, when we remove the 4 Ω resistor, it effectively becomes 2-loop circuit.

Then I calculated current *I* using loop-method on these two loops, and ended up with 4.4 A. After that I calculated Thevenin voltage by subtracting voltage loss of leftmost 10 Ω resistor (4.4 A * 10 Ω - 90 = 46 V) and subtracted the effect of 20 V voltage source, ending up with Vth = 26 V.

26 V / (4.77 Ω Rth + 4 Ω from load resistor) = 2.965 A == Correct

So yes, this approach would never work effectively with very complicated circuits (10+ loops). I had to do it this way since the exercise is due tomorrow and I think this is the method I was expected to use. But like I said, the simplifying method shown here is something I must take another look at. Thanks to both of you for help and hints so far.